Resultados da pesquisa

Dx\right)\right\}\vdash \left(Ax\land Cx\right)} { ∀ x ( A x → B x ) , A l } ⊢ ∃ x B x {\displaystyle \left\{\forall x\left(Ax\to Bx\right),Al\right\}\vdash...

\left(\left(A\to B\right)\to \left(A\to C\right)\right)} ⊢ ( D → ( B → A ) ) → ( B → ( D → A ) ) {\displaystyle \vdash \left(D\to \left(B\to A\right)\right)\to \left(B\to...

a\left\{x-x_{1}\right\}\left[x-x_{2}\right]=0} a ( x − x 1 ) ( x − x 2 ) = 0 {\displaystyle a\left(x-x_{1}\right)\left(x-x_{2}\right)=0} então, por fim...

M x ) , H s } ⊢ M s {\displaystyle \left\{\forall x\left(Hx\to Mx\right),Hs\right\}\vdash Ms} CUIDADO Lembre-se que esta regra é aplicável a fórmulas...

{sen} \,a\right)\;\!,} B ( cos ⁡ ( a + b ) , s e n ( a + b ) ) {\displaystyle B\left(\cos \left(a+b\right),\mathrm {sen} \,\left(a+b\right)\right)\;\!} e...

R\right)\to \left(P\to \left(Q\to R\right)\right)\right)\to \left(\left(Q\to R\right)\to \left(\left(P\to Q\right)\to \left(P\to R\right)\right)\right)}...

0}{\frac {Q_{V}}{\Delta T}}\;=\;\left({\frac {\partial E}{\partial T}}\right)_{V}} a volume constante C P = lim Δ T → 0 Q P Δ T = ( ∂ H ∂ T ) P {\displaystyle...

(y)=2\cdot \mathrm {sen} \,\left({\frac {x+y}{2}}\right)\mathrm {cos} \,\left({\frac {x-y}{2}}\right)} s e n ( x ) − s e n ( y ) = 2 ⋅ s e n ( x − y 2...

\left\{\left(A\land B\right)\to C,\neg C\right\}\vdash \neg A\lor \neg B} ¬ ( A ∧ ¬ B ) ⊢ A → B {\displaystyle \neg \left(A\land \neg B\right)\vdash A\to B}...

dE\;=\;T\left[\left({\frac {\partial S}{\partial V}}\right)_{T}dV\;+\;\left({\frac {\partial S}{\partial T}}\right)_{V}dT\right]\;-PdV} ou ainda: d E = [ T ( ∂ S ∂ V...

H}{\partial P}}\right)_{T}\;=\;\left({\frac {\partial (E+PV)}{\partial P}}\right)_{T}\;=\;\left({\frac {\partial E}{\partial P}}\right)_{T}\;+\;\left({\frac...

\left(B\lor C\right),\neg \neg B\right\}\vdash A\land B} { A ∨ B , B → C , ¬ A ∧ D } ⊢ C {\displaystyle \left\{A\lor B,B\to C,\neg A\land D\right\}\vdash C}...

∨ ( ¬ ¬ C ∨ D ) {\displaystyle \therefore B\lor \left(\neg \neg C\lor D\right)} As tabelas de verdade não parecem muito práticas neste caso. Afinal, temos...

{\displaystyle \left(A\to B\right)\equiv \neg \left(A\land \neg B\right)\equiv \left(\neg A\lor B\right)\equiv \neg \left(\neg A\downarrow B\right)} Da mesma forma:...

{\partial \left({\frac {G}{T}}\right)}{\partial T}}\right)_{P}\;=\;{\frac {1}{T}}\left({\frac {\partial G}{\partial T}}\right)\;-\;{\frac {G}{T^{2}}}} As...

\left(u\;+\;{\frac {v^{2}}{2}}\right)\delta _{V}\right)\;+\;\left(-\;\left[\rho _{1}\left(u_{1}\;+\;{\frac {v_{1}^{2}}{2}}\right)\;-\;\sigma _{1}\right](v_{1}A_{1})\;+\;\left[\rho...

dH\;=\;T\left[\left({\frac {\partial S}{\partial P}}\right)_{T}dP\;+\;\left({\frac {\partial S}{\partial T}}\right)_{V}dT\right]\;+\;VdP} d H = T [ − ( ∂ V ∂ T ) P d...

{\displaystyle \left|\delta w\right|\;=\;\left|fdl\right|\;=\;\left|P_{externo}-Adl\right|\;=\;\left|P_{externo}dV\right|} Porém , se este volume aumentar...

{\partial ^{2}v_{x}}{\partial z^{2}}}\right)\;+\;0\;=\;\rho _{0}\left(0\;+\;0\;+\;0\;+\;0\right)} μ 0 ∂ 2 v x ∂ z 2 = ∂ p ∂ x {\displaystyle \mu...

{r_{1}}{r_{2}}}\right)^{2}\right)^{2}}}\left[\ln \left({\frac {r_{2}}{r_{1}}}\right)\;+\;{\frac {1}{2}}\;\left(\left({\frac {r_{1}}{r_{2}}}\right)^{2}\;-\;1\right)\right]}...